浙江大学PAT_甲级_1019. General Palindromic Number (20)

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A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number N > 0 in base b >= 2, where it is written in standard notation with k+1 digits a_i as the sum of (a_ibⁱ) for i from 0 to k. Here, as usual, 0 <= a_i < b for all i and a_k is non-zero. Then N is palindromic if and only if a_i = a_k-i for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any non-negative decimal integer N and a base b, you are supposed to tell if N is a palindromic number in base b.

Input Specification:

Each input file contains one test case. Each case consists of two non-negative numbers N and b, where 0 <= N <= 10⁹ is the decimal number and 2 <= b <= 10⁹ is the base. The numbers are separated by a space.

Output Specification:

For each test case, first print in one line "Yes" if N is a palindromic number in base b, or "No" if not. Then in the next line, print N as the number in base b in the form "a_k a_k-1 ... a₀". Notice that there must be no extra space at the end of output.

Sample Input 1:

27 2

Sample Output 1:

Yes
1 1 0 1 1

Sample Input 2:

121 5

Sample Output 2:

No
4 4 1

我的C++c程序：

#include<iostream>
#include<vector>
#include<stack>
using namespace std;
bool check(vector<int> vec)//判断vector里的数是否为回文，是返回1
{
	int left = 0, right = vec.size() - 1;
	while (left <= right)
	{
		if (vec[left] != vec[right])
		{
			return 0;
		}
		left++;
		right--;
	}
	return 1;
}
vector<int> transform(long long n, long long d)//将10进制数转换为D进制，并将各位存储到vector里
{
	stack<int>sta;
	vector<int>vec;
	while (n != 0)
	{
		sta.push(n%d);
		n = n / d;
	}
	while (!sta.empty())
	{
		vec.push_back(sta.top());
		sta.pop();
	}
	return vec;//返回一个vector
}
int main()
{
	vector<int> vec;
	long long n, d;//n一个数，d转换后的进制
	cin >> n >> d;
	if (n == 0||n==1||n==2)//0 1 2在任何进制都是回文数
	{
		cout << "Yes"<<endl;
		cout <<n;
		return 0;
	}
	vec = transform(n, d);//进制转换
	if (check(vec))//检查回文数
		cout << "Yes" << endl;
	else
		cout << "No" << endl;
	for (int i = 0;i < vec.size()-1;i++)//输出回文数
	{
		cout << vec[i]<<' ';
	}
	cout << vec[vec.size() - 1];//输出回文数最后一位
	//system("pause");
	return 0;
}